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Exercise 1.
Exam Mean
Writing 56.4
Software use 68.7
Standard Deviation
9.3
5.6
Bill
Ted
65
70
67
75
There were no questions about the above data, so I’m ignoring it…
You are asked to do a study of shelters for abused and battered women to determine the
necessary capacity in your city to provide housing for most of these women. After
recording data for a whole year, you find that the mean number of women in shelters
each night is 250, with a standard deviation of 75. Fortunately, the distribution of the
number of women in the shelters each night is normal, so you can answer the following
questions posed by the city council.
A. If a city’s shelters have a capacity of 350, will that be enough place for abused
women on 95% of all nights? If not, what number of shelter openings will be
needed?
The z value for the top 95% from a table is:
1.645
Use the z formula:
z = (x-mu)/sigma
1.645 = (x-250)/75
x = 75*1.645+250
x = 373.375
Round up to x = 374
The capacity is not enough. They need 374 spots.
B. The current capacity is only 220 openings, because some shelters have been
closed. What is the percentage of nights that the number of abused women
seeking shelter will exceed current capacity?
Get the z score:
z = (x-mu)/sigma
z = (220-250)/75
z = -0.4
prob(z > -0.4) from a table is:
0.6554
= 65.54% of nights
Second assignment
10. For the total population of a large southern city, mean family income is $34,000, with
a standard deviation (for the population) of $5,000.
A. Imagine that you are taking a subsample of 200 city residents. What is the
probability that your sample mean is between $33,000 and $34,000?
z(33000) = (33000-34000)/(5000/sqrt(200))
z = -2.8284
z(34000) = 0, since it’s the mean
prob(-2.8284 < z < 0) from a table is:
0.4977
B. For this same sample size, what is the probability that the same mean exceeds
$37,000?
z(37000) = (37000-34000)/(5000/sqrt(200))
Z = 8.4853
prob(z > 8.4853) = 0 (it’s just slightly above 0, but any table shows 0)
Third assignment
PRESTG80
Occupational
Prestige Score
Statistic
High School Diploma
Bachelor’s Degree
Mean
Median
Std. Deviation
Minimum
Range
Interquartile Range
Mean
Median
Std. Deviation
Minimum
Maximum
Range
Interquartile Range
40.50
40.00
11.142
17
74
74
52.74
51.00
12.675
17
86
69
20
A. Construct the 90% confidence interval for occupational prestige for respondents
with only a high school diploma (N=2,153).
The z value for 90% confidence is 1.645.
The interval goes from:
mean-z*sd/sqrt(N) to mean+z*sd/sqrt(N)
40.5-1.645*11.142/sqrt(2153) to 40.5+1.645*11.142/sqrt(2153)
40.10 to 40.90
B. Construct the 90% confidence interval for occupational prestige for respondents
with a bachelor’s degree (N=750). State in words the meaning of the result.
The z value for 90% confidence is 1.645.
The interval goes from:
mean-z*sd/sqrt(N) to mean+z*sd/sqrt(N)
52.74-1.645*12.675/sqrt(750) to 52.74+1.645*12.675/sqrt(750)
51.98 to 53.50
We can be 90% confident that the true mean prestige value for these people lies
between these values.
C. Use these statistics to discuss differences in occupation prestige scores by
educational attainment.
The entire confidence interval for bachelor’s degrees is higher than the interval for
just high school diplomas. Therefore, we can be very confident that there is indeed a
higher prestige for people with a degree.
Fourth assignment
1. A subsample of the 2006 MTF survey suggests that adolescents are generally
concerned with social issues. In fact, 70% of the 1,482 respondents who answered the
question reported that they either sometimes or often think about social issues.
Estimate at the 95% and 99% confidence levels the proportion of all adolescents who
sometimes or often think about social issues.
95%
The z value is 1.96.
The interval goes from:
p-z*sqrt(p*(1-p)/N) to p+z*sqrt(p*(1-p)/N)
0.7-1.96*sqrt(0.7*0.3/1482) to 0.7+1.96*sqrt(0.7*0.3/1482)
0.677 to 0.723
99%
The z value is 2.5758.
The interval goes from:
p-z*sqrt(p*(1-p)/N) to p+z*sqrt(p*(1-p)/N)
0.7-2.5758*sqrt(0.7*0.3/1482) to 0.7+2.5758*sqrt(0.7*0.3/1482)
0.669 to 0.731